The sales pitch is this: There exists an algorithm that can determine whether any first-order statement about the reals involving polynomial equations and inequalities is true or false. The precise statement is more technical, but that’s the idea^{1}. Apparently we’ve known that this algorithm has existed since the 1960’s, and have had an effective implementation since the 1970’s. Well, it was news to me!
There are lots of silly examples I could give, but let me just dive into a real, recent problem I was thinking about.
For which integers $a$ and $b$ does the cubic
\[\begin{align*} f(s, t) = 1 - 2 a s &+ a^2 s^2 - b s^2 + s^3 + a b s^3 - b t \\ &- 3 s t + a b s t + a s^2 t + b^2 s^2 t + a t^2 + 2 b s t^2 + t^3 \end{align*}\]have a positive minimum on the region $\{a s + bt \geq 1\} \cap [0, 1]^2$?
This is an annoying problem. I solved it using computer algebra to automate my calculus-based analysis, but I didn’t get an answer I was happy with. CAD can do it almost automatically!
We can phrase our question like this: For which $a, b$ does the formula
\[\forall s, t\ [0 \leq s, t \leq 1,\ as + bt \geq 1 \implies f(s, t) > 0]\]hold? (Note that we don’t need another variable for the minimum itself. The region is compact, so $f(s, t) > 0$ is enough.) If we give this statement to a CAD algorithm, then it will turn it into an unquantified statement, meaning that it only depends on $a$ and $b$. Here’s what Mathematica’s built in CAD does:
f = 1 - 2 a s + a^2 s^2 - b s^2 + s^3 + a b s^3 - b t - 3 s t + a b s t + a s^2 t + b^2 s^2 t + a t^2 + 2 b s t^2 + t^3
CylindricalDecomposition[
ForAll[{s, t},
Implies[0 <= s <= 1 && 0 <= t <= 1 && a*s + b*t >= 1, f > 0]],
{b, a}]
(b <= -(17/4) && (a < (2 - b)/2 - 1/2 Sqrt[-4 + b^2] || a > Root[27 - 4 b^3 + 18 b #1 - b^2 #1^2 + 4 #1^3 &, 3]))
|| (-(17/4) < b <= -2 && (a < (2 - b)/2 - 1/2 Sqrt[-4 + b^2] || a > (2 - b)/2 + 1/2 Sqrt[-4 + b^2]))
|| -2 < b < 1
|| (1 <= b < 2.08... && a > Root[27 - 4 b^3 + 18 b #1 - b^2 #1^2 + 4 #1^3 &, 1])
|| (b == 2.08... && a > (2 - b)/2 + 1/2 Sqrt[-4 + b^2])
|| (b > 2.08... && a > Root[27 - 4 b^3 + 18 b #1 - b^2 #1^2 + 4 #1^3 &, 1])
The output is a disjoint union of conditions which, together, are equivalent to our input. The full output is a bit messy, so here it is again with only the parts relevant to “integers $b \geq 1$”:
(1 <= b < 2.08... && a > Root[27 - 4 b^3 + 18 b #1 - b^2 #1^2 + 4 #1^3 &, 1])
|| (b > 2.08... && a > Root[27 - 4 b^3 + 18 b #1 - b^2 #1^2 + 4 #1^3 &, 1])
And now, noticing that the inner condition on $a$ is the same in both clauses, and that we only care about integer values, we see that this whole thing is equivalent to just this:
a > Root[27 - 4 b^3 + 18 b #1 - b^2 #1^2 + 4 #1^3 &, 1]
This says that our original statement about $f$ is equivalent to “$a$ is larger than the smallest root $x$ of the cubic $27 - 4 b^3 + 18 b x - b^2 x^2 + 4 x^3$.” This sounds unhelpful, but it is actually quite strong.
Using CAD again, we can determine when this cubic has only one real root. Because it’s a cubic, this is the negation of the statement
\[\exists x < y < z \ [q(x) = q(y) = q(z) = 0],\]where $q$ is the cubic in question. Let’s pass that to Mathematica:
q = 27 - 4 b^3 + 18 b x - b^2 x^2 + 4 x^3
CylindricalDecomposition[
Exists[{x, y, z},
x < y < z &&
q == 0 && (q /. {x -> y}) == 0 && (x /. {x -> z}) == 0], {b}]
b < 1.89...
If $b > 1$ (we only care about integers), then the cubic only has one root. In that case, because its leading coefficient in $x$ is positive, being larger than the real root is equivalent to saying that the cubic is positive. If $b = 1$, then we can just do CAD again but substitute $b = 1$ first, which reveals that the condition is $a > -1$.
So, all together, we have this: If $b = 1$, then $f(s, t) > 0$ on the region for all positive integers $a$; if $b \geq 2$, then $f(s, t) > 0$ on the region if and only if
\[27 + 4 a^3 + 18 a b - a^2 b^2 - 4 b^3 > 0.\]This is the very precise, quantitative statement I wanted all along, and it only took a few minutes of computing time (plus some more exploring) to discover. Amazing!
And even this pitch lowballs CAD. If you get into the underlying geometry, what CAD computes tells you more than whether particular first-order statements are true. ↩
Resolve to be honest at all events; and if in your own judgment you cannot be an honest lawyer, resolve to be honest without being a lawyer. Choose some other occupation, rather than one in the choosing of which you do, in advance, consent to be a knave.
Bill Clinton took C-SPAN on a tour of the Oval Office and told them that he had been relearning geometry to help his daughter in school. Nixon famously had an audio recording hobby. (That one produced lots of candid White House conversations.)
On the 80th anniversary of D-Day, it seems fitting to recall a truly remarkable collection of “mundane” presidential remarks: Franklin Delano Roosevelt’s fireside chats. These were relaxed radio broadcasts where FDR laid out his opinions, motivations, and policy decisions in simple language to the American people. There are transcripts of every address and recordings of quite a few. While some of the content is dry and outdated (can you really call the gold standard gripping?), the collection as a whole offers a fascinating window into FDR’s decision making.
And FDR, the greatest anti-democratic tyrant of all time, had a lot to talk about. Bank runs (it turns out the gold standard is a bit gripping), the New Deal, the Supreme Court, and obviously the war. Even 90 years later, many themes from these chats are worth revisiting.
In ten or twenty years, we should quiz highschool students on whether the following quote is FDR talking about the Axis powers, or Biden talking about Russia:
It is easy for you and for me to shrug our shoulders and to say that conflicts taking place thousands of miles from the continental United States, and, indeed, thousands of miles from the whole American Hemisphere, do not seriously affect the Americas — and that all the United States has to do is to ignore them and go about its own business. Passionately though we may desire detachment, we are forced to realize that every word that comes through the air, every ship that sails the sea, every battle that is fought does affect the American future.
Of course once you listen to the speech you will know immediately that it is FDR, because he is one of the most powerful orators the United States has ever had, and Biden is Biden^{1}.
While Russia continues its invasion of Ukraine, Biden argues with Congress about sending military aid, NATO conducts exercises aimed at countering Russian aggression, and Putin now suggests that he could target European countries with long range missiles. FDR called Hawaii “an outpost of defense in the Pacific,” and there is a good argument that Taiwan is the same. Is the world now so dissimilar to what FDR saw when he looked across the oceans?
One thing you learn from the chats is that FDR had a remarkable talent for framing. Look at how he begins his first fireside chat after the United States entered the war:
This is not a fireside chat on war. It is a talk on national security, because the nub of the whole purpose of your President is to keep you now, and your children later, and your grandchildren much later, out of a last-ditch war for the preservation of American independence and all of the things that American independence means to you and to me and to ours.
Really masterful stuff!
FDR was famously unhappy with the Supreme Court. Conservative Justices, appointed by earlier presidents, struck down many of his economic policies, igniting debate about how much the court truly represented and served the people. Does this sound familiar to you?^{2}. FDR threatened to appoint new, younger Justices which would agree with him more often. And this was not some backroom, wink-wink threat; he went on the radio and said this:
It is the American people themselves who are in the driver’s seat.
It is the American people themselves who want the furrow plowed.
It is the American people themselves who expect the third horse to pull in unison with the other two.
…
What is my proposal? It is simply this: whenever a Judge or Justice of any Federal Court has reached the age of seventy and does not avail himself of the opportunity to retire on a pension, a new member shall be appointed by the President then in office, with the approval, as required by the Constitution, of the Senate of the United States.
Congress eventually decided this was too far and backed out, but the threat was serious enough that the Supreme Court coincidentally started agreeing with FDR more. You can see FDR’s incredible framing at work again in this fireside chat:
If by that phrase “packing the Court” it is charged that I wish to place on the bench spineless puppets who would disregard the law and would decide specific cases as I wished them to be decided, I make this answer: that no President fit for his office would appoint, and no Senate of honorable men fit for their office would confirm, that kind of appointees to the Supreme Court.
Something that I find frustrating about many public political discussions is that they are very “guarded.” Politicians try to not look bad, so they ignore certain topics or distort well-known facts. I understand that this is strategically necessary to some extent, but sometimes it annoys me.
FDR’s willingness to get on the radio and plainly explain himself was an invaluable public service for voters and later generations, but also a politically brilliant move. We should expect more of this from our public servants. (And maybe, as FDR suggested, we should be harsher to the Supreme Court.)
I hope that, should very awful things happen in the next few decades, as happened in FDR’s lifetime, we have a president that can speak with the strength and clarity that FDR did in his fireside chats. So hopefully it doesn’t happen in the next four years!^{3}
With respect to the 80th D-Day anniversary again, it is due to the great efforts and sacrifices of the men and women of the Allied forces that we and the free people of Europe and Japan do get to expect such things from our public servants. So, on perhaps the last big anniversary where any of them will be with us, I offer a very sincere thank you to these heroes.
]]>“How can it equal one?” he said. “If one times one equals one that means that two is of no value because one times itself has no effect. One times one equals two because the square root of four is two, so what’s the square root of two? Should be one, but we’re told it’s two, and that cannot be.”
I was reminded of this episode by Howard’s recent appearance on Joe Rogan’s podcast, where he was billed as a “researcher in the fields of logic and engineering,” among other things. Obviously his reasoning about $1 \times 1$ is rough around the edges, but can you blame a man with no formal training for being sloppy? Howard took a few years to refine his thinking and then posted a proof that $1 \times 1 = 2$ using his system of “Terryology.” It starts like this (edited for clarity):
First and foremost let us ask the most obvious question, is this a finished equation? Yes, or No? The answer is, No! Let us start our forensic audit there.
It is an incomplete equation. Why? Because it’s not even on both sides. Yes, Nature desires action but demands equilibrium. Therefore, in order for an equation to be finished / completed both sides of the equation must be equally, balanced.
First of all, this is very funny. But second of all, I really admire Howard’s interest in basic arithmetic that many take for granted. It’s easy to justify $2 \times 5 = 10$ when you treat multiplication as a shortcut for addition, as in
\[2 \times 5 = 2 + 2 + 2 + 2 + 2,\]but then what do things like $2 \times \frac{1}{5}$ or $\sqrt{3} \times \sqrt{2}$ or $e \times \pi$ mean? How do we know what multiplication actually does? How can we be sure that we’re doing the operations “correctly”?
These questions about arithmetic are surprisingly difficult to answer if you want to be rigorous about them. As an exercise, go define what $x^y$ means for arbitrary reals $x$ and $y$, then prove that this is well-defined and satisfies all the usual properties. (No, you aren’t allowed to assume that logarithms exist.) If you’ve done it properly, I’d bet that you would struggle to explain it to a layman.
So I really, genuinely, like Howard’s curiosity. I will go a step further and say that Howard is right as far as multiplication is concerned. He has been unfairly attacked from within so-called “mainstream mathematics” for his unorthodox views, but his views are completely correct when viewed with the right perspective. For example, in his proof that $1 \times 1 = 2$, he states this:
Associate and Commutative Law: “When (a) and (b) are positive integers, [then] (a) is to be added to itself as many times as there are units in (b).
Putting aside that this is not what either the associative or commutative law says (again, no formal training!), Howard is giving a rule to evaluate $a \times b$ when both are positive integers. He thinks it should be
\[a \times b = a + \underbrace{(a + a + \cdots + a)}_{b\text{ times}}.\]Of course the traditional, mainstream view of multiplication would tell us that
\[a \times b = \underbrace{(a + a + \cdots + a)}_{b\text{ times}}\]but this is merely a difference in notation. If we take the view that mathematicians using $\tau = 2\pi$ are brave freedom fighters struggling against oppressive orthodoxy, as some in our profession do, then why should the movement end there? Should we be surprised to learn that more sophisticated, tasteful mathematicians of the future use the Howard multiplication operator
\[a \cdot_H b = a \times (b + 1)?\]I would say not.
If we are lucky, the day will come when identities such as $1 \cdot_H 1 = 2$ are viewed as trivial and obvious the same way that the “identity” $1 \times 1 = 1$ is today. On that day, we will look back on Howard’s writing as we now look back on the surviving scraps of mathematics from Euclid’s time. Deep, prescient, and insightful. Amen.
(I finished my PhD last week, so everything here is said as a Doctor in Mathematics with the full confidence invested in me by Rutgers, The State University of New Jersey, and the graduate faculty therein.)
]]>I would like to wish everyone, including all haters and losers (of which, sadly, there are many) a truly happy and enjoyable Memorial Day!
— Donald J. Trump (@realDonaldTrump) May 24, 2015My thesis, which I defended last Monday, includes a big list of acknowledgements. Unlike the very stately, presidential message above, it does not include nods to haters or losers. It does include a few unusual entities, like the United States and the Coca-Cola Foundation, but everything remains positive.
Even taking a pretty expansive view of who deserved an acknowledgement, I still made some omissions for space or style. Below is a list of entities, thanked without constraints.
Service workers. Everyday, something in my life is easier because of cleaning staff, restaurant workers, delivery drivers, or other service workers. These men and women work hard, menial jobs and often get very little thanks or recognition (even less than administrative staff like secretaries). For example, the Rutgers math department cleaning staff often works at night, when you can’t see them! Thank you to all the workers who make life a thousand times easier for all of us.
Euclid et al. Although “mathematics” consists of many people working constantly to push our boundary of knowledge forward, there are some people that pushed this boundary further than anyone else. The mathematical community owes a great debt to these legendary mathematicians of yore. Some of the “standard” mentions are Euler, Gauss, Newton, and Leibniz, but I feel that a special nod must be given to Euclid for being the first to shape mathematics into its modern form.
Taxpayers. About 66% of all academics work at public universities. Even outside of public universities, many grants are awarded by NSF, NSA, or other federal agencies, and students often pay tuition using federal loans. Maybe Princeton and Harvard could get by with private money, but the majority of academics work at places that can’t. It is thanks to the United States and her citizens that I and most academics have work. A grateful thank you to them.
Tech wizards. It is unthinkable to do mathematics without the computer tools I use every day. Leslie Lamport and Don Knuth made $\LaTeX$, the symbolic computation group at the University of Waterloo created Maple, and Guido van Rossum designed the almost-perfect Python. I must give a special thank you to the late Bram Moolenaar, who created the truly stupendous text editor Vim. I do 90% of my work by writing notes in Vim, and that work is all the more pleasant and efficient thanks to him.
That covers just about everyone. At this point, if you haven’t been thanked, I have some bad news for you. If it’s any consolation, I’m not dying, so you still have a chance to do something really nice to make the next list. Better start now!
]]>If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many also fall in continued proportion between the numbers which have the same ratios with the original numbers.
If you have trouble seeing $\sqrt{2}$ here, don’t worry; so does everyone who reads the Elements. (Professor David Joyce, explaining how to parse this, remarks “If 1 is considered to be a number, the argument simplifies.”) With all due respect to Euclid, in hindsight we know that mediating all of mathematics through lines and circles makes some things more difficult than they need to be.
Things are better today now that we have symbolic notation, though on occasion people fall into a Euclid-style trap and say that mathematics is symbol crunching. If you can move the right funny letters around in calculus, linear algebra, or differential equations, then you’re a “good student.” But of course math is not just symbol crunching, just like it isn’t just lines and circles.
This historical outline guides me when thinking about AI generated art. There has been a lot of uproar over the past few years from concerned artists about AI tools like Midjourney. Are these tools going to take jobs from already-struggling artists? Are they just stealing original work and repackaging it with fancy post-processing? And whatever you think, can’t we agree that “AI art” is not real art?
This sentiment that “AI art is not real art” seems mistaken. I understand that creating art is very difficult, in the same way that drawing and reasoning about complicated geometric diagrams is difficult. But are the techniques used to create art today what art is? Can it really be said that if you aren’t drawing lines on a tablet, or putting paint on a canvas, you aren’t making art? I don’t buy it.
I suspect that artists lashing out at AI art are more concerned about job security than they are artistic ideals. And fair enough. AI art might not be taking their jobs now, but it probably will. Just look at Cosmo’s cover from 2022, or read the story of Kelly McKernan, whose distinctive art style was reproduced by an AI startup that McKernan is suing for copyright infringement. It is demoralizing to devote a chunk of your life to something only for a newfangled machine to do it better. And yet, like the old folk hero John Henry, you can fight progress your whole life and accomplish nothing but dying.
AI art does not need to be the end of human art. If we, as a society, think that human art is important, then we can fund it. We already do fund it in various ways, like through the National Endowment for the Arts^{1}. I would be happy to see more money go that way. We fund lots of things that no one would ever pay for on their own^{2}, and art is one of the more noble things we could pick.
It’s great that I can ask ChatGPT to explain things, and it’s great that more people can make art using tools like Midjourney. If you are worried about a future where these tools are so widespread that AI does everything, a future where your economic opportunities are extremely limited, a future where humanity is left to passively observe a digital world, then you better start writing blog posts saying nice things about ChatGPT before it takes over.
NEA has a very neat map where you can see how much money they put into individual states. ↩
Like most of academia. Does New Jersey need 100 math Ph.D students at its flagship university? No, of course not. But it’s a nice, fairly cheap thing that brings in smart, motivated people and might pay off in unexpected ways. I’m very grateful to citizens of the Garden State for their generous support! ↩
I want more smart people to go into public service, and fewer smart people doing weird, illegal finance things. So, below are two brief examples of good government that I hope might inspire some plucky students to consider becoming bureaucrats.
The Supplemental Nutrition Assistance Program (SNAP), also known as food stamps, is a federal program that gives low-income Americans money for food. It feeds millions of Americans who might otherwise go hungry.
A great part of SNAP is that it’s dirt cheap. It seems untoward, but cost is an important part of welfare programs. How many people need money? How much will we give them? How much will this improve their standards of living? Helping people is good, but there’s a big difference between a billion dollars and a trillion dollars. SNAP comes in at under three percent of the annual federal budget.
We knew that SNAP would be affordable thanks to estimates from the Congressional Budget Office (CBO), whose economists and policy analysts model costs and budgetary effects. CBO models run on information from the Department of Agriculture (USDA), which administers SNAP, the Census Bureau, which collects demographic information, and the Federal Reserve, which monitors economic conditions.
The planning and analysis of SNAP implicates a handful of federal agencies, but the day to day operations are handled by states themselves, which gives us more than fifty agencies and probably thousands of employees. This sounds like a vast bureaucracy, but employees are very cheap! The total cost of all federal employees in 2019 was around 300 billion dollars, which is roughly the cost of SNAP itself.
For a tiny portion of annual spending, the federal government gets an army of dedicated public servants who provide quality data, analysis, and oversight related to of a wide array of programs like SNAP. This is an excellent and often overlooked feature of the federal government.
In 1957, nine black Americans enrolled in Little Rock Central High School. They had been barred from attending by segregation, but the Supreme Court had just ruled that this was unconstitutional^{1}. In a shocking move, Arkansas governor Orval Faubus defied the Supreme Court and ordered the Arkansas National Guard to forcibly block the students from entering school.
The federal government obviously has to intervene here. You can’t have governors using pseudo-armies to defy court orders. It isn’t as feel-good as feeding hungry people, but an important function of government is maintaining order and enforcing laws. Fortunately, President Eisenhower understood this and sent in the Army.
Whenever normal agencies prove inadequate to the task and it becomes necessary for the Executive Branch of the Federal Government to use its powers and authority to uphold Federal Courts, the President’s responsibility is inescapable.
In accordance with that responsibility, I have today issued an Executive Order directing the use of troops under Federal authority to aid in the execution of Federal law at Little Rock, Arkansas. This became necessary when my Proclamation of yesterday was not observed, and the obstruction of justice still continues.
—President Eisenhower, 1957
Eisenhower’s ability to do this requires a lot of moving parts.
First, there needs to be a federal law enforcement mechanism which can force states to do things. That’s a professional military. Second, there needs to be some legitimate authority for the President to use force to restore order. Congress provided this in the Insurrection Act of 1807. Finally, the President needs to have the character to use the tools at his disposal wisely. We can’t have the military shooting pickpockets in New York City.
It’s an old American pastime to complain about Congress and the president, but they are public servants who make important legislative and executive decisions. Had Congress not recognized the importance of a strong military, if they had been more squeamish about enforcing laws, or if Eisenhower had been a little racist himself, school integration could have played out very differently.
Government is necessary, but good government takes good public servants. We need more smart people in every branch, every agency, and every office. Math students tend to not have the charisma necessary to be politicians^{2}, but there are a lot of other good options!
There are openings every summer at the Department of Energy, the NSA always wants mathematicians (but has some questionable hiring practices), and you definitely qualify for jobs in the 1500 series. This includes jobs at Census, IRS, GAO, NIST, Treasury, and more. If you are interested in the judiciary, then you will be comforted to know that math students tend to do pretty well on the LSAT.
I would encourage all math students to at least consider an internship in government. You can spare one summer to try it out!
The Court also upheld segregation years earlier, but American segregation was happening before the Court ever said a word about it. ↩
Though there was Ira Glasser, so you never know. ↩
The basic idea is that we want to prove the identity
\[\begin{equation} \label{goal} \sum_{k = 0}^{n - 1} {6n \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{6n - k} = \sum_{k = 0}^{n - 1} R(k) {6k \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{5k} \end{equation}\]where $R(k)$ is some explicit rational function. The reason is that the sum on the right has no $n$ in it, and after seeing $R(k)$ written out, is obviously monotonic because the terms are positive. The sum on the left contains a bunch of $n$’s, so it isn’t obvious if it gets bigger or smaller as $n$ grows.
First, a quick lesson on WZ theory. A sequence $a(n, k)$ is called hypergeometric provided that $a(n + 1, k) / a(n, k)$ and $a(n, k + 1) / a(n, k)$ are both explicit rational functions in $n$ and $k$. Certain nice hypergeometric sequences have a “partner” $b(n, k)$ which satisfies the identity
\[a(n + 1, k) - a(n, k) = b(n, k + 1) - b(n, k).\]In this case $(a, b)$ is called a WZ pair. It is simple to algorithmically determine whether any explicit sequence is hypergeometric or part of a WZ pair. Our summand happens to be both.
Say that you have a WZ pair $(a, b)$. That is,
\[\begin{equation} \label{wz} a(n + 1, k) - a(n, k) = b(n, k + 1) - b(n, k). \end{equation}\]Let
\[S(n) = \sum_{k = 0}^{n - 1} a(n, k)\]be the sum that we are interested in. Summing over \eqref{wz} from $k = 0$ to $k = n$ gives
\[S(n + 1) - S(n) - a(n, n) = b(n, n + 1) - b(n, 0).\]If we move $a(n, n)$ to the right-hand side, then the left-hand side is telescoping, so summing again recovers $S(n)$ itself. The end-result will be
\[S(n) = \sum_k (a(k, k) + b(k, k + 1) - b(k, 0)),\]where the bounds are too tedious to determine. In our case it turns out that $b(k, 0) = 0$, and by some standard theory about hypergeometric sequences $b(k, k + 1)$ is a rational multiples of $a(k, k)$, so we can write
\[S(n) = \sum_k R(k) a(k, k),\]the kind of sum we desired.
The proof now is merely a matter of executing the steps outlined in the
previous section. All of these have been implemented in Maple or
Mathematica for decades. If we download
EKHAD and read it
into Maple, then executing zeillim(a(n, k), k, n, N, 0, 1)
produces output
that implies \eqref{goal}. The rational function is
The age distribution of Austrians walking around cities skewed much older than what I would expect in America. It was not uncommon for most people on a tram or sitting at a cafe to be in their forties, fifties, or higher. Young people were around but only made up majorities in the city center or at night. This seems different than what you would see in New York City, Atlanta, DC, or even my small hometown in South Carolina.
It turns out that Austrians are measurably older than Americans. In the US, about 1 in 3 people (36%) are 50 or older. In Austria, the number is about 2 in 5 (42%). In total, Austria’s population is the 14th oldest in the world, while the US is 62nd. This doesn’t completely explain what I saw (there are regional differences in age distributions, plus I was subject to confirmation bias), but it makes me feel a little less crazy.
I’d be concerned if I was Austrian. Aging populations mean fewer workers support more people, so there’s less money to go around for everyone. Austria could try to get more workers, but their fertility rates have fallen to around 1.4 children per woman, the country seems to have a reputation for being unfriendly to immigrants and raising the retirement age makes people angry. They haven’t had much success improving their slow productivity growth, either.
Austria isn’t alone with its demographic problems. Most developed countries are in the same boat, which the New York Times described in an excellent write-up earlier this year. The key takeaways are that Europe and East Asia are aging very fast, and the US is aging slower thanks to high immigration. The only places getting younger in the near future are Africa and India.
This is a great time to make it easier to move to and stay in the US. Far more people want to move here than we accept. Our current immigration policy is to make most of them to wait, possibly for decades. Why not let them in?
The Cato Institute has a collection of proposals for how to increase immigration. One that I like, from the perspective of deal-making, is expanding visa access for high-skilled workers. Republicans don’t want to take the huddled masses yearning to be free, but maybe they’ll agree to take the educated masses yearning to work at Google^{1}.
At a bare minimum, we should get back to normal, pre-COVID levels of immigration. We handed out embarassingly few green cards during the pandemic. Let’s open the doors again.
Did you know that the US has a special tax treaty with India that lets Indian students in the US take the standard deduction? A non-Indian foreign graduate student pays around a thousand dollars a year more in taxes than Indians or Americans. ↩
Take, for example, power generation. Every factory needs power. One very efficient late game technology is nuclear power, which requires the following supply chain:
The catch is that step 2 is probabilistic. For each unit of uranium ore that you process, there is a 99.3% chance to obtain uranium-238 and a 0.7% chance to obtain uranium-235. Let’s call these “bad” uranium and “good” uranium, respectively. Step 3 requires that you have 10 bad uranium and 1 good uranium. These probabilities make the setup for nuclear power very slow.
Things liven up once you generate 40 good uranium. Then, via another late-game technology, you can put 40 good uranium into a machine that spits out 41 good uranium. This is 41 total good uranium—the 40 you put in plus one new one. It doesn’t sounds like much but multiplying your supply by $41 / 40$ forever would lead to exponential growth.
Unfortunately, this is not how things work. The process to turn 40 good uranium into 41 good uranium only works on 40 uranium at a time per machine you have to run it. The first time you run it you will have 41 good uranium, but you can only put exactly 40 back into the process. The lone new uranium sits there, useless.
This brings us to the neat question. If you had infinitely many machines to run this process, then the amount of good uranium at step $n$ is given recursively by
\[\begin{align*} U_0 &= 40 \\ U_{n + 1} &= U_n + \left \lfloor \frac{U_n}{40} \right \rfloor = \left \lfloor \frac{41}{40} U_n \right \rfloor. \end{align*}\]What is the asymptotic value of $U_n$?
A good idea is to ignore the floor and guess that the answer is close to $(41 / 40)^n \cdot 40$ anyway. This is wrong, but it’s not that wrong.
$n$ | $U_n$ | $(41 / 40)^n \cdot 40$ | ratio |
---|---|---|---|
0 | 40 | 40 | 1 |
1 | 41 | 41 | 1 |
2 | 42 | 42 | 1 |
40 | 80 | 107 | 0.745 |
100 | 291 | 472 | 0.616 |
Like any good question, you can find a version of this in the OEIS. If, instead of $41 / 40$, you use $3 / 2$, then you will find A061418. This led me to a paper of Odlyzko and Wilf studying similar sequences.
Adapting the ideas in Wilf’s paper, it turns out that $U_n$ is exponential but smaller than $(41 / 40)^n \cdot 40$. There is a constant $c$ such that
\[U_n \sim (41 / 40)^n 40 c \quad ; \quad c = 0.574256\dots\]So you lose about 43% of your exponential capacity to waiting, but this is still pretty good!
The constant $c$ does not seem to be expressible in terms of anything well-known. There is a “formula,” but it is useless for practical purposes.
Unsurprisingly, Wilf and Odlyzko were not considering uranium processing in Factorio. They were working on a solution to the Josephus problem.
The Josephus problem is this: Arrange $n$ people in a circle and execute every $q$th person until there is only one left. Where should you stand in the initial circle to be the surviving person? This position is called $J_q(n)$. The $q = 2$ case has the famous solution $J_2(n) = 2(n - 2^{\lfloor \log_2 n \rfloor}) + 1$, but there aren’t closed forms beyond that.
Concrete Mathematics has a neat algorithm to solve this problem. Once you pass over someone, move their index to the smallest index not yet seen. (If $q = 3$ then $1$ will become $n + 1$ and $2$ will become $n + 2$.) The last person to be executed will have the label $qn$, so you just need to determine the original index of the person who is eventually labeled $qn$. The sequences of Wilf and Odlyzko help do this.
Our sequence helps solve a similar Josephus-type problem. Arrange $n$ people in a circle and execute every $q$th person starting from the first one. That is, you will always execute the first person, then count off by $q$ from there. Under the same relabeling scheme as before, the last person to be executed will have label $q(n - 1) + 1$. It turns out that the label of this person after going in “reverse” $k$ steps is $qn - U_k^{(q)}$, where $U_k^{(q)}$ is defined by
\[\begin{align*} U_0^{(q)} &= q - 1 \\ U_k^{(q)} &= \left\lfloor \frac{q}{q - 1} U_{k - 1}^{(q)} \right\rfloor. \end{align*}\]This sequence solves the problem because we can stop computing as soon as $qn - U_k^{(q)} \leq n$. That is, as soon as $U_k^{(q)} \geq (q - 1)n$.
This sequence is also exactly the same form as our uranium sequence if we set $q = 41$. That is, when we are producing good uranium in Factorio, we are implicitly solving this modified Josephus problem where we execute every forty-first person. If you have $U$ good uranium the first time that you cross $40n$, then then the survivor stood at initial position $41n - U$.
This is not quite proving that Factorio is Turing complete (it is), but it is a neat fact to notice. Factorio has more mathematical depth than any game I’ve played before. I hope that this one example demonstrates that.
]]>This year I won two of the TA teaching awards given by the Rutgers math department to graduate students. In the past 25 years the department has given 100 such awards, but only five people have ever won more than one, and none more than two. I don’t know if this is a Joe Pesci, “Thank you” moment, or a Sally Fields, “You like me!” moment. Either way, I am glad to be in the company of many excellent teachers^{1}.
I am particularly happy to contribute to the excellence of teaching in my academic bloodline. I wrote a small script to gather data on the recipients of teaching assistant awards, and am pleased to share the results.
Which advisor at Rutgers produces students who win the most teaching assistant awards? Doron Zeilberger.
Advisor^{2} | Students | TA Excellence awards |
---|---|---|
Zeilberger | 21 | 11 |
Huang | 12 | 5 |
Woodward | 9 | 4 |
Tunnell | 6 | 4 |
Lepowsky | 13 | 4 |
Of course, Professor Zeilberger has many students, which gives him more chances to get lucky. To account for this, take advisors with at least five students and sort by awards per student. Zeilberger is still second.
Advisor | Students | TA Excellence awards per student |
---|---|---|
Tunnell | 6 | 0.666667 |
Zeilberger | 21 | 0.52381 |
Woodward | 9 | 0.444444 |
Huang | 12 | 0.416667 |
Wilson | 5 | 0.4 |
To really get a feel for this data you should just look at the picture.
While Zeilberger is very far away from the rest of the faculty, this scatter plot does not convey how far away. Almost all advisors “produce” zero TA teaching awards.
Even if you think that the TA Excellence awards are flukes, according to the department only five math graduate students have ever won the more prestigious School of Arts and Sciences Award for Distinguished Contributions to Undergraduate Education. Four of them—Andrew Baxter, Lara Pudwell, Paul Raff, and Matthew Russell—are Zeilberger students.
It does not surprise me that Dr. Z gets the best teachers. Here is the first line of Opinion 77:
We professional academic mathematicians are lucky to be paid for something we enjoy doing the most, mathematical research. Many of us also enjoy teaching, and I for one, would be willing to pay, if I had to, for the privilege to teach, even Freshman calculus.
Whether Dr. Z puts this ethos into his students, or just happens to attract ones that already have it, I don’t know. But there is obviously something in the water, so to speak.
It has been a real joy to teach all of my students over the past four years. Every thankful email I get, every nice (or constructive!) comment in evaluations, every time someone (unwisely) asks me for a letter of recommendation, I feel happy that I’m helping someone along. If these awards reflect anything, they reflect that.
Each award comes with \$200. I’ve donated \$100 to the OEIS and $100 to the Wikimedia foundation. ↩
This data was collected by matching teaching award recipients with their listed advisors on the graduation page. For current students who had not graduated, I attempted to determine their advisor myself. This worked for all but three current students.
The data also only counts students who existed after the creation of the TA Excellence award. For example, Tunnel had 7 students in total, but one was before the award existed. ↩